Point $C$ is on the segment $AB$ which has endpoints $A(-1, 0)$  and $B(3, 8)$.  Point $C$ is three times as far from point $A$ as it is from  point $B$.  What are the coordinates of point $C$?
We are told that $AC = 3CB$, so $AB = AC + CB = 4CB$.  Let $M$ be the midpoint of $\overline{AB}$. Then, we have $BM = \dfrac{AB}{2}$.

Since $AB = 4CB$, we have $CB = \dfrac{AB}{4} = \dfrac{BM}{2}$.  In other words, $C$ is the midpoint of $\overline{BM}$.

Since $M$ is the midpoint of $\overline{AB}$, we have $M = \left(\dfrac{-1+3}{2} , \dfrac{0+8}{2}\right) = (1,4)$.

Similarly, since $C$ is the midpoint of $\overline{BM}$, we have $C = \left(\dfrac{3 + 1}{2}, \dfrac{8 + 4}{2}\right) = \boxed{(2,6)}$.